📂Electrodynamics

Delayed Potential on Continuous Distribution of Delay Times

Overview¹

The scalar and vector potentials for a moving point charge are called retarded potentials, and they are as follows.

$$ \begin{align*} V(\mathbf{r},\ t) &= \dfrac{1}{4\pi\epsilon_{0}} \int \dfrac{ \rho (\mathbf{r}^{\prime},\ t_{r}) }{ \cR } d\tau^{\prime} \\[1em] \mathbf{A}( \mathbf{r},\ t) &= \dfrac{\mu_{0}}{4\pi} \int \dfrac{\mathbf{J}(\mathbf{r}^{\prime},\ t_{r})}{\cR}d\tau^{\prime} \end{align*} $$

Here, $t_{r}$ is the retarded time.

Retarded Time

If the charge and current distribution do not change over time, the scalar and vector potentials satisfy the following Poisson’s equations.

$$ \nabla^2 V=-\dfrac{1}{\epsilon_{0}} \rho,\quad \nabla^2 \mathbf{A}=-\mu_{0}\mathbf{J} $$

Solving these yields:

$$ \begin{equation} V(\mathbf{r})=\dfrac{1}{4\pi\epsilon_{0}} \int \dfrac{ \rho (\mathbf{r}^{\prime}) }{ \cR } d\tau^{\prime},\quad \mathbf{A}( \mathbf{r} ) = \dfrac{\mu_{0}}{4\pi} \int \dfrac{\mathbf{J}(\mathbf{r}^{\prime})}{\cR}d\tau^{\prime} \end{equation} $$

$\bcR$ is the separation vector.

However, since electromagnetic waves propagate at the speed of light, if the current distribution moves, the “current moment’s” potential is not due to the current distribution but rather due to the distribution at “some past moment”. It’s important that it is not the current distribution that matters, but the distribution from “some past moment” when the electromagnetic wave started. The distance traveled is $\cR$, and with the speed of light being $c$, the moment the electromagnetic wave started, reaching the current moment $\mathbf{r}$, is as follows.

$$ t_{r} \equiv t-\dfrac{\cR}{c} $$

This is called the retarded time. To put it more simply, the news arriving at “the current moment” $t$ started at the time corresponding to the retarded time. It is expressed as a moment in time, rather than time itself, for this reason. When dealing with a moving point charge or current distribution, the separation vector is not $\mathbf{r}-\mathbf{r}^{\prime}$, but $\bcR=\mathbf{r}-\mathbf{w}$.

Retarded Potentials

Thus, when the charge distribution changes over time, in other words, when the charge moves, the generalization of $(1)$ is as follows.

$$ V(\mathbf{r},\ t)=\dfrac{1}{4\pi\epsilon_{0}} \int \dfrac{ \rho (\mathbf{r}^{\prime},\ t_{r}) }{ \cR } d\tau^{\prime},\quad \mathbf{A}( \mathbf{r},\ t) = \dfrac{\mu_{0}}{4\pi} \int \dfrac{\mathbf{J}(\mathbf{r}^{\prime},\ t_{r})}{\cR}d\tau^{\prime} $$

$\rho (\mathbf{r}^{\prime}, t_{r})$ is the charge density at point $\mathbf{r}^{\prime}$ when the time is $t_{r}$. The two potentials described above are called retarded potentials. These equations were not mathematically derived. Instead, they were reasonably explained with physically correct logic. Although there is a logical leap, fortunately, the results match reality well. To prove this, it must be checked whether the newly obtained potentials satisfy the following wave equation $(2)$ and the Lorentz gauge $(3)$.

$$ \begin{equation} \begin{aligned} \Box ^2 V &= -\dfrac{1}{\epsilon_{0}}\rho \\ \Box ^2 \mathbf{A} &= -\mu_{0}\mathbf{J} \end{aligned} \end{equation} $$

$$ \begin{equation} \nabla \cdot \mathbf{A} = -\mu_{0} \epsilon_{0} \frac{\partial V}{\partial t} \end{equation} $$

This thorough verification is necessary because applying the same logic to electric and magnetic fields, instead of potentials, results in outcomes that do not match reality.